Basis functions and positive definiteness












0












$begingroup$


Suppose I have a set of functions $g_i(x)$ which form a basis on some interval $[a,b]$. My question is whether the matrix
$$
A_{ij} = int_a^b g_i(x) g_j(x) dx
$$
is positive definite?



My approach so far is to consider a function $f: [a,b] rightarrow mathbb{R}$,
$$
f(x) = sum_i c_i g_i(x) = c^T g(x)
$$
Then we have:
begin{align}
0 &le int_a^b |f(x)|^2 dx \
& = int_a^b c^T g(x) g(x)^T c dx \
&= c^T left( int_a^b g(x) g(x)^T dx right) c \
&= c^T A c
end{align}
This shows that $A$ is postive semi-definite.



But I wonder if it is correct to argue that $int_a^b |f(x)|^2 dx$ is equal to $0$ only if $c = 0$, because otherwise if $c ne 0$, then that implies that the $g_i(x) = 0$ for all $i$ and all $x$, contradicting the assumption that $g_i(x)$ is a basis. Is that correct reasoning?



As a followup question, consider the matrix
$$
B_{ij} = int_a^b g'_i(x) g'_j(x) dx
$$
This matrix is positive semi-definite by the same reasoning above, but is it also positive definite? If the argument above is correct, I don't think it applies to the matrix $B$, because if $g_i(x)$ forms a basis on $[a,b]$, that doesn't necessarily mean that $g'_i(x)$ is also a basis?



So the question is whether $A$ and/or $B$ are positive definite matrices?






linear-algebra matrices







share|cite|improve this question









$endgroup$












  • $begingroup$
    I have edited my answer and included a counterexample for $B$.
    $endgroup$
    – Kavi Rama Murthy
    Jan 18 at 12:47
















0












$begingroup$


Suppose I have a set of functions $g_i(x)$ which form a basis on some interval $[a,b]$. My question is whether the matrix
$$
A_{ij} = int_a^b g_i(x) g_j(x) dx
$$
is positive definite?



My approach so far is to consider a function $f: [a,b] rightarrow mathbb{R}$,
$$
f(x) = sum_i c_i g_i(x) = c^T g(x)
$$
Then we have:
begin{align}
0 &le int_a^b |f(x)|^2 dx \
& = int_a^b c^T g(x) g(x)^T c dx \
&= c^T left( int_a^b g(x) g(x)^T dx right) c \
&= c^T A c
end{align}
This shows that $A$ is postive semi-definite.



But I wonder if it is correct to argue that $int_a^b |f(x)|^2 dx$ is equal to $0$ only if $c = 0$, because otherwise if $c ne 0$, then that implies that the $g_i(x) = 0$ for all $i$ and all $x$, contradicting the assumption that $g_i(x)$ is a basis. Is that correct reasoning?



As a followup question, consider the matrix
$$
B_{ij} = int_a^b g'_i(x) g'_j(x) dx
$$
This matrix is positive semi-definite by the same reasoning above, but is it also positive definite? If the argument above is correct, I don't think it applies to the matrix $B$, because if $g_i(x)$ forms a basis on $[a,b]$, that doesn't necessarily mean that $g'_i(x)$ is also a basis?



So the question is whether $A$ and/or $B$ are positive definite matrices?






linear-algebra matrices







share|cite|improve this question









$endgroup$












  • $begingroup$
    I have edited my answer and included a counterexample for $B$.
    $endgroup$
    – Kavi Rama Murthy
    Jan 18 at 12:47














0












0








0





$begingroup$


Suppose I have a set of functions $g_i(x)$ which form a basis on some interval $[a,b]$. My question is whether the matrix
$$
A_{ij} = int_a^b g_i(x) g_j(x) dx
$$
is positive definite?



My approach so far is to consider a function $f: [a,b] rightarrow mathbb{R}$,
$$
f(x) = sum_i c_i g_i(x) = c^T g(x)
$$
Then we have:
begin{align}
0 &le int_a^b |f(x)|^2 dx \
& = int_a^b c^T g(x) g(x)^T c dx \
&= c^T left( int_a^b g(x) g(x)^T dx right) c \
&= c^T A c
end{align}
This shows that $A$ is postive semi-definite.



But I wonder if it is correct to argue that $int_a^b |f(x)|^2 dx$ is equal to $0$ only if $c = 0$, because otherwise if $c ne 0$, then that implies that the $g_i(x) = 0$ for all $i$ and all $x$, contradicting the assumption that $g_i(x)$ is a basis. Is that correct reasoning?



As a followup question, consider the matrix
$$
B_{ij} = int_a^b g'_i(x) g'_j(x) dx
$$
This matrix is positive semi-definite by the same reasoning above, but is it also positive definite? If the argument above is correct, I don't think it applies to the matrix $B$, because if $g_i(x)$ forms a basis on $[a,b]$, that doesn't necessarily mean that $g'_i(x)$ is also a basis?



So the question is whether $A$ and/or $B$ are positive definite matrices?






linear-algebra matrices







share|cite|improve this question









$endgroup$




Suppose I have a set of functions $g_i(x)$ which form a basis on some interval $[a,b]$. My question is whether the matrix
$$
A_{ij} = int_a^b g_i(x) g_j(x) dx
$$
is positive definite?



My approach so far is to consider a function $f: [a,b] rightarrow mathbb{R}$,
$$
f(x) = sum_i c_i g_i(x) = c^T g(x)
$$
Then we have:
begin{align}
0 &le int_a^b |f(x)|^2 dx \
& = int_a^b c^T g(x) g(x)^T c dx \
&= c^T left( int_a^b g(x) g(x)^T dx right) c \
&= c^T A c
end{align}
This shows that $A$ is postive semi-definite.



But I wonder if it is correct to argue that $int_a^b |f(x)|^2 dx$ is equal to $0$ only if $c = 0$, because otherwise if $c ne 0$, then that implies that the $g_i(x) = 0$ for all $i$ and all $x$, contradicting the assumption that $g_i(x)$ is a basis. Is that correct reasoning?



As a followup question, consider the matrix
$$
B_{ij} = int_a^b g'_i(x) g'_j(x) dx
$$
This matrix is positive semi-definite by the same reasoning above, but is it also positive definite? If the argument above is correct, I don't think it applies to the matrix $B$, because if $g_i(x)$ forms a basis on $[a,b]$, that doesn't necessarily mean that $g'_i(x)$ is also a basis?



So the question is whether $A$ and/or $B$ are positive definite matrices?






linear-algebra matrices




linear-algebra matrices






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 18 at 7:12









vibevibe

1648




1648












  • $begingroup$
    I have edited my answer and included a counterexample for $B$.
    $endgroup$
    – Kavi Rama Murthy
    Jan 18 at 12:47


















  • $begingroup$
    I have edited my answer and included a counterexample for $B$.
    $endgroup$
    – Kavi Rama Murthy
    Jan 18 at 12:47
















$begingroup$
I have edited my answer and included a counterexample for $B$.
$endgroup$
– Kavi Rama Murthy
Jan 18 at 12:47




$begingroup$
I have edited my answer and included a counterexample for $B$.
$endgroup$
– Kavi Rama Murthy
Jan 18 at 12:47










1 Answer
1






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oldest

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$begingroup$

$sum c_ic_j int_a^{b} g_i'(x)g_j'(x), dx=int_a^{b}sum c_ic_j int_a^{b} g_i'(x)g_j'(x), dx=int_a^{b} |sum c_ig_i'(x)|^{2}, dx geq 0$. So $B$ is non-negative definite. But $sum c_ic_j int_a^{b} g_i'(x)g_j'(x), dx=0$ only implies $sum c_ig_i=c$ for some constant $c$ and $B$ need not be positive definite. For a counterexample consider $g_1(x)=x,g_2(x)=1-x$ on $[0,1]$. It is easy to see that $B$ is not positive definite.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you, this makes sense
    $endgroup$
    – vibe
    Jan 18 at 15:37











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$begingroup$

$sum c_ic_j int_a^{b} g_i'(x)g_j'(x), dx=int_a^{b}sum c_ic_j int_a^{b} g_i'(x)g_j'(x), dx=int_a^{b} |sum c_ig_i'(x)|^{2}, dx geq 0$. So $B$ is non-negative definite. But $sum c_ic_j int_a^{b} g_i'(x)g_j'(x), dx=0$ only implies $sum c_ig_i=c$ for some constant $c$ and $B$ need not be positive definite. For a counterexample consider $g_1(x)=x,g_2(x)=1-x$ on $[0,1]$. It is easy to see that $B$ is not positive definite.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you, this makes sense
    $endgroup$
    – vibe
    Jan 18 at 15:37
















0












$begingroup$

$sum c_ic_j int_a^{b} g_i'(x)g_j'(x), dx=int_a^{b}sum c_ic_j int_a^{b} g_i'(x)g_j'(x), dx=int_a^{b} |sum c_ig_i'(x)|^{2}, dx geq 0$. So $B$ is non-negative definite. But $sum c_ic_j int_a^{b} g_i'(x)g_j'(x), dx=0$ only implies $sum c_ig_i=c$ for some constant $c$ and $B$ need not be positive definite. For a counterexample consider $g_1(x)=x,g_2(x)=1-x$ on $[0,1]$. It is easy to see that $B$ is not positive definite.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you, this makes sense
    $endgroup$
    – vibe
    Jan 18 at 15:37














0












0








0





$begingroup$

$sum c_ic_j int_a^{b} g_i'(x)g_j'(x), dx=int_a^{b}sum c_ic_j int_a^{b} g_i'(x)g_j'(x), dx=int_a^{b} |sum c_ig_i'(x)|^{2}, dx geq 0$. So $B$ is non-negative definite. But $sum c_ic_j int_a^{b} g_i'(x)g_j'(x), dx=0$ only implies $sum c_ig_i=c$ for some constant $c$ and $B$ need not be positive definite. For a counterexample consider $g_1(x)=x,g_2(x)=1-x$ on $[0,1]$. It is easy to see that $B$ is not positive definite.






share|cite|improve this answer











$endgroup$



$sum c_ic_j int_a^{b} g_i'(x)g_j'(x), dx=int_a^{b}sum c_ic_j int_a^{b} g_i'(x)g_j'(x), dx=int_a^{b} |sum c_ig_i'(x)|^{2}, dx geq 0$. So $B$ is non-negative definite. But $sum c_ic_j int_a^{b} g_i'(x)g_j'(x), dx=0$ only implies $sum c_ig_i=c$ for some constant $c$ and $B$ need not be positive definite. For a counterexample consider $g_1(x)=x,g_2(x)=1-x$ on $[0,1]$. It is easy to see that $B$ is not positive definite.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 18 at 12:46

























answered Jan 18 at 7:16









Kavi Rama MurthyKavi Rama Murthy

62.9k42362




62.9k42362












  • $begingroup$
    Thank you, this makes sense
    $endgroup$
    – vibe
    Jan 18 at 15:37


















  • $begingroup$
    Thank you, this makes sense
    $endgroup$
    – vibe
    Jan 18 at 15:37
















$begingroup$
Thank you, this makes sense
$endgroup$
– vibe
Jan 18 at 15:37




$begingroup$
Thank you, this makes sense
$endgroup$
– vibe
Jan 18 at 15:37


















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